What is the integral from 0 to 1 of x^4e^{4x} ?

The integral from 0 to 1 of x^4e^{4x} is (15e^4-3)/(128)

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integral from 0 to 1 of x^4e^{4x}

\int_{0}^{1} x^{4} e^{4 x} d x

\frac{15 e ^{4} - 3}{128}

Decimal

6.37478 \dots

Solution steps

\int_{0}^{1} x^{4} e^{4 x} d x

Apply Integration By Parts

= [\frac{1}{4} e^{4 x} x^{4} - \int e^{4 x} x^{3} d x]_{0}^{1}

\int e^{4 x} x^{3} d x = \frac{1}{256} (64 e^{4 x} x^{3} - 3 (16 e^{4 x} x^{2} - 2 (4 e^{4 x} x - e^{4 x})))

= [\frac{1}{4} e^{4 x} x^{4} - \frac{1}{256} (64 e^{4 x} x^{3} - 3 (16 e^{4 x} x^{2} - 2 (4 e^{4 x} x - e^{4 x})))]_{0}^{1}

Simplify

[\frac{1}{4} e^{4 x} x^{4} - \frac{1}{256} (64 e^{4 x} x^{3} - 3 (16 e^{4 x} x^{2} - 2 (4 e^{4 x} x - e^{4 x})))]_{0}^{1} : [\frac{1}{128} e^{4 x} (32 x^{4} - 32 x^{3} + 24 x^{2} - 12 x + 3)]_{0}^{1}

= [\frac{1}{128} e^{4 x} (32 x^{4} - 32 x^{3} + 24 x^{2} - 12 x + 3)]_{0}^{1}

Compute the boundaries:

\frac{15 e ^{4}}{128} - \frac{3}{128}

= \frac{15 e ^{4}}{128} - \frac{3}{128}

Simplify

= \frac{15 e ^{4} - 3}{128}

What is the integral from 0 to 1 of x^4e^{4x} ?
The integral from 0 to 1 of x^4e^{4x} is (15e^4-3)/(128)